wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

100 mL of a 0.1 M CH3COONa(aq) is titrated against a 0.1 M HCl(aq). Calculate the pH of the solution when 50 mL of HCl was added.
Given: pKa(CH3COOH)=4.74 at 25oC

A
6.21
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
5.85
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
5.24
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
4.74
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D 4.74
mmol of CH3COONa=100×0.1=10
mmol of HCl=50×0.1=5

CH3COONa (aq)+HCl (aq)CH3COOH (aq)+NaCl (l)Initial: 10 5 0 0Final: 5 0 5 5

After titration equal milimoles of CH3COOH and CH3COONa will form. The resultant solution will acts as an acidic buffer.
Final concentration=Moles Total Volume
Total Volume =100+50=150 mL
[CH3COOH]final=5150 M
[CH3COONa]final=5150 M
The formula to calculate the pH of an acidic buffer is :
pH=pKa+log([CH3COONa][CH3COOH])pH=4.74+log51505150pH=4.74

flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Hydrolysis
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon